# Henderson-Hasselbalch approximation

The Henderson-Hasselbalch approximation gives us a method to approximate the pH of a buffer solution. The basic equation is as follows:

$$\mathrm{pH} \approx \mathrm{p}K_\mathrm{a} + \log\dfrac{\ce{[A^-]}}{\ce{[HA]}}$$

Note that HA is a weak acid. By definition, "HA does not dissociate completely and we can say $$\ce{[HA]}\approx\ce{[HA]_i}$$ and $$\ce{[A−]}\approx\ce{[A−]_i}$$. Hence, we can use the initial concentrations because $$\mathrm{p}K_\mathrm{a} + \log\dfrac{\ce{[A^-]_i}}{\ce{[HA]_i}} \approx \mathrm{p}K_\mathrm{a} + \log\dfrac{\ce{[A^-]}}{\ce{[H\!A]}} = \mathrm{pH}$$

Source: Chem.Libretexts

Now they have given a question and it's solution as:

Find $$\ce{[H+]}$$ in a solution $$\pu{1.0 M}$$ $$\ce{HNO2}$$ and $$\pu{0.225 M}$$ $$\ce{NaNO2}$$ The $$K_\mathrm{a}$$ for $$\ce{HNO2}$$ is $$5.6\times10^{−4}$$.

$$\mathrm{pH} = 3.14 + \log \left( \dfrac{1}{0.225} \right)$$

So my question is why did they neglect the concenteration of conjugate base from the dissociation of the weak acid and included concentration of the base from the salt?

• 1) Your last equation is wrong. 2) The actual error is small (2.60 vs 2.62). 3) At these concentrations, you would need to account for ionic strength. Depending of what you want to do, it depends of how accurate you want to be or not. Mar 27 '18 at 1:43